Problem: Evaluate the triple integral. $ \int_1^2 \int_x^2 \int_0^y y - z \, dz \, dy \, dx =$
We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_1^2 \int_x^2 \int_0^y y - z \, dz \, dy \, dx \\ \\ &= \int_1^2 \int_x^2 \left[ yz - \dfrac{z^2}{2} \right]_0^y dy \, dx \\ \\ &= \int_1^2 \int_x^2 y^2 - \dfrac{y^2}{2} \, dy \, dx \\ \\ &= \int_1^2 \int_x^2 \dfrac{y^2}{2} \, dy \, dx \end{aligned}$ The second layer: $\begin{aligned} &\int_1^2 \int_x^2 \dfrac{y^2}{2} \, dy \, dx \\ \\ &= \int_1^2 \left[ \dfrac{y^3}{6} \right]_x^2 dx \\ \\ &= \int_1^2 \dfrac{4}{3} - \dfrac{x^3}{6} \, dx \end{aligned}$ The third layer: $\begin{aligned} &\int_1^2 \dfrac{4}{3} - \dfrac{x^3}{6} \, dx \\ \\ &= \left[ \dfrac{4x}{3} - \dfrac{x^4}{24} \right]_1^2 \\ \\ &= \left( \dfrac{8}{3} - \dfrac{16}{24} \right) - \left( \dfrac{4}{3} - \dfrac{1}{24} \right) \\ \\ &= \dfrac{2}{3} + \dfrac{1}{24} \\ \\ &= \dfrac{17}{24} \end{aligned}$ In conclusion: $ \int_1^2 \int_x^2 \int_0^y y - z \, dz \, dy \, dx = \dfrac{17}{24}$